Chemistry Calculator

Calculate OH Concentration and pH of 10 M NaCN

Use this interactive sodium cyanide hydrolysis calculator to determine hydroxide ion concentration, pOH, pH, cyanide hydrolysis extent, and related equilibrium values. The default setup is for 10.0 M NaCN at 25 degrees Celsius, using the acid dissociation constant of HCN to compute the base dissociation constant of CN^–.

NaCN pH Calculator

NaCN concentration

Concentration unit

Temperature assumption

Ka of HCN

Solution method

Enter values and click Calculate to see OH concentration and pH for NaCN.

How to Calculate OH Concentration and pH of 10 M NaCN

When you need to calculate OH concentration and pH of 10 M NaCN, you are solving a classic weak base hydrolysis problem. Sodium cyanide, NaCN, is a highly soluble ionic compound that dissociates essentially completely in water into Na⁺ and CN^–. The sodium ion is a spectator ion for acid-base chemistry, but cyanide is the conjugate base of hydrogen cyanide, HCN, which is a weak acid. That means CN^– reacts with water to produce hydroxide ions:

CN^– + H₂O ⇌ HCN + OH^–

The key result is that a sodium cyanide solution is basic, not neutral. In a concentrated solution such as 10 M NaCN, the hydroxide concentration is determined by the base hydrolysis equilibrium rather than by simply assuming complete conversion to OH^–. This is a very important distinction in general chemistry, analytical chemistry, and industrial chemical calculations.

Step 1: Identify the relevant equilibrium

The cyanide ion is a weak base, so the correct equilibrium expression is the base dissociation constant:

K_b = [HCN][OH^–] / [CN^–]

Most textbooks and reference tables provide the acid dissociation constant of HCN instead of the base dissociation constant for CN^–. At 25 degrees C, a commonly used value is:

K_a for HCN ≈ 6.2 × 10^-10
K_w = 1.0 × 10^-14

Because K_a × K_b = K_w, we get:

K_b = K_w / K_a = (1.0 × 10^-14) / (6.2 × 10^-10) ≈ 1.61 × 10^-5

Step 2: Set up an ICE table for 10 M NaCN

Since NaCN dissociates fully, the initial cyanide concentration is approximately 10.0 M. Let x represent the amount of CN^– that reacts with water:

Initial: [CN^–] = 10.0, [HCN] = 0, [OH^–] = 0
Change: [CN^–] = -x, [HCN] = +x, [OH^–] = +x
Equilibrium: [CN^–] = 10.0 – x, [HCN] = x, [OH^–] = x

Substitute into the K_b expression:

1.61 × 10^-5 = x² / (10.0 – x)

Step 3: Solve for hydroxide concentration

For a weak base where x is small relative to the starting concentration, the approximation 10.0 – x ≈ 10.0 is excellent:

x² = (1.61 × 10^-5)(10.0) = 1.61 × 10^-4

x = √(1.61 × 10^-4) ≈ 0.0127 M

Therefore:

[OH^–] ≈ 1.27 × 10^-2 M
[HCN] ≈ 1.27 × 10^-2 M
[CN^–] at equilibrium ≈ 9.9873 M

If you solve the quadratic exactly, the answer is essentially the same at this concentration and K_b value. That is why the approximation is widely accepted for classroom and practical calculations alike.

Quick answer: For 10 M NaCN at 25 degrees C using K_a(HCN) = 6.2 × 10^-10, the calculated hydroxide concentration is about 0.0127 M, the pOH is about 1.90, and the pH is about 12.10.

Step 4: Convert OH concentration to pOH and pH

Once [OH^–] is known, the rest is straightforward:

pOH = -log[OH^–]
pH = 14.00 – pOH at 25 degrees C

Using [OH^–] = 0.0127 M:

pOH = -log(0.0127) ≈ 1.90
pH = 14.00 – 1.90 ≈ 12.10

Why NaCN produces a basic solution

This behavior is explained by conjugate acid-base theory. HCN is a weak acid, so its conjugate base, CN^–, has measurable basicity. In water, CN^– removes a proton from H₂O, creating HCN and OH^–. The stronger the conjugate base, the more the equilibrium shifts toward hydroxide production. Cyanide is not as strong a base as hydroxide itself, but it is strong enough to produce an appreciably high pH, especially at a very large concentration like 10 M.

Approximate versus exact solution

In many introductory problems, instructors expect the square root approximation:

x = √(K_bC)

This is valid when x is much smaller than the initial concentration C. For 10 M NaCN:

x ≈ 0.0127 M
x / C ≈ 0.0127 / 10 = 0.00127 = 0.127%

Since the hydrolysis is far below 5%, the approximation is exceptionally safe. The exact quadratic is:

x² + K_bx – K_bC = 0

Solving:

x = [-K_b + √(K_b² + 4K_bC)] / 2

Plugging in the numbers changes the answer only by a tiny amount beyond the significant figures usually reported in chemistry problems.

Comparison table: NaCN concentration versus calculated alkalinity

NaCN concentration (M)	Approx. [OH-] (M)	pOH	pH at 25 degrees C	Percent hydrolyzed
0.010	4.01 × 10^-4	3.40	10.60	4.01%
0.10	1.27 × 10^-3	2.90	11.10	1.27%
1.0	4.01 × 10^-3	2.40	11.60	0.401%
10.0	1.27 × 10^-2	1.90	12.10	0.127%

The trend is clear: as cyanide concentration rises by a factor of 10, the hydroxide concentration increases by roughly the square root of 10 because of the weak base relation x = √(K_bC). The pH rises steadily, but the fraction hydrolyzed actually decreases because the denominator becomes much larger.

Reference constants and data you should know

Different data tables may list slightly different values for K_a of HCN, usually in the neighborhood of 4.9 × 10^-10 to 6.2 × 10^-10 depending on source, ionic strength treatment, and rounding conventions. That means your final pH can shift slightly by a few hundredths of a pH unit. For educational problem solving, this is normal and expected.

Parameter	Typical value	Why it matters
K_w at 25 degrees C	1.0 × 10^-14	Needed to convert K_a to K_b
K_a of HCN	6.2 × 10^-10	Controls cyanide basicity in water
K_b of CN^–	1.61 × 10^-5	Used directly in hydrolysis calculations
Computed [OH^–] for 10 M NaCN	0.0127 M	Main result for this problem
Computed pH for 10 M NaCN	12.10	Shows the solution is strongly basic

Common mistakes when solving this problem

Assuming NaCN is neutral because it is a salt. Not all salts are neutral. Salts of weak acids and strong bases are basic.
Using K_a directly instead of converting to K_b. The hydrolyzing species is CN^–, so base equilibrium is required.
Setting [OH^–] equal to 10 M. That would only be appropriate for a strong base like NaOH, not for cyanide hydrolysis.
Forgetting to compute pOH before pH. Once [OH^–] is known, pOH comes first.
Ignoring significant figures and reference constant variation. Small differences in K_a can slightly shift the final pH.

Real chemistry considerations for concentrated solutions

A 10 M solution is very concentrated. In advanced chemistry, such concentrated ionic solutions may deviate from ideal behavior because activities are not identical to concentrations. That means a rigorous treatment could require activity coefficients rather than simple molarity alone. However, for most educational calculations, exam problems, and quick estimation workflows, the standard equilibrium concentration method remains the accepted approach. The calculator above follows that conventional method.

Safety context and practical significance

Sodium cyanide is an extremely hazardous chemical used in some industrial contexts such as metal extraction and plating. Its aqueous chemistry matters because pH strongly affects cyanide speciation between CN^– and HCN. At lower pH values, more hydrogen cyanide can form, and HCN is highly toxic and volatile. This is one reason why maintaining alkaline conditions is a critical safety practice in cyanide handling systems.

If you are studying cyanide chemistry, it is worth reviewing safety and toxicology information from authoritative agencies. The following sources are useful:

Worked summary for the exact question

Start with 10.0 M NaCN, so [CN^–]_initial = 10.0 M.
Use K_a(HCN) = 6.2 × 10^-10.
Compute K_b(CN^–) = 1.0 × 10^-14 / 6.2 × 10^-10 = 1.61 × 10^-5.
Solve x² / (10.0 – x) = 1.61 × 10^-5.
Obtain x ≈ 0.0127 M = [OH^–].
Calculate pOH = 1.90.
Calculate pH = 12.10.

Final answer

The hydroxide ion concentration for a 10 M NaCN solution is approximately 1.27 × 10^-2 M. The corresponding pOH is about 1.90 and the pH is about 12.10 at 25 degrees C, assuming K_a(HCN) = 6.2 × 10^-10 and ideal solution behavior.

Calculate Oh Concentration And Ph Of 10 M Nacn